altitude H; or in more general terms the product of the base of a cylinder by its altitude cannot be the measure of a less cylinder.

We say, in the second place, that this same product cannot be the measure of a greater cylinder; for, not to multiply figures, let CD be the radius of the base of the given cylinder; and, if it be possible, let surf. CD x H be the measure of a greater cylinder, of a cylinder, for example, of which CA is the radius of the base and H the altitude.

The same construction being supposed as in the first case, the prism circumscribed about the given cylinder will have for its measure GHIP × H; the area GHIP is greater than surf. CD; consequently, the solidity of the prism in question is greater than surf. CD × H; the prism then would be greater than the cylinder of the same altitude whose base is surf. CA. But the prism on the contrary is less than the cylinder, since it is contained in it; therefore it is impossible that the product of the base of a cylinder by its altitude should be the measure of a greater cylinder.

We conclude then, that the solidity of a cylinder is equal to the product of its base by its altitude.

517. Corollary 1. Cylinders of the same altitude are to each other as their bases, and cylinders of the same base are to each other as their altitudes.

518. Corollary 11. Similar cylinders are to each other as the cubes of their altitudes, or as the cubes of the diameters of the bases. For the bases are as the squares of their diameters; and, since the cylinders are similar, the diameters of the bases are as the altitudes (510); consequently the bases are as the squares of the altitudes; therefore the bases multiplied by the altitudes, or the cylinders themselves, are as the cubes of the altitudes.

519. Scholium. Let R be the radius of the base of a cylinder, H its altitude, the surface of the base will be л R2 (291), and the solidity of the cylinder will be л R2 × Н, oг л R' H.

LEMMA.

520. The convex surface of a right prism is equal to the perimeter of its base multiplied by its altitude.

Demonstration. This surface is equal to the sum of the rect

angles AFGB, BGHC, CHID, &c. (fig. 252), which compose it. Fig. 252.

Now the altitudes AF, BG, CH, &c., of these rectangles are each equal to the altitude of the prism. Therefore the sum of these rectangles, or the convex surface of the prism, is equal to the perimeter of its base multiplied by its-altitude.

521. Corollary. If two right prisms have the same altitude, the convex surfaces of these prisms will be to each other as the perimeters of the bases.

LEMMA.

522. The convex surface of a cylinder is greater than the convex surface of any inscribed prism, and less than the convex surface of any circumscribed prism.

Demonstration. The convex surface of the cylinder and that of Fig 252. the inscribed prism ABCDEF (fig. 252) may be considered as having the same length, since every section made in the one and the other parallel to AF is equal to AF; and if, in order to obtain the magnitude of these surfaces, we suppose them to be cut by planes parallel to the base, or perpendicular to the edge AF, the sections will be equal, the one to the circumference of the base, and the other to the perimeter of the polygon ABCDE less than this circumference; since therefore, the lengths being equal, the breadth of the cylindric surface is greater than that of the prismatic surface, it follows that the first surface is greater than the second.

By a course of reasoning entirely similar it may be shown that the convex surface of the cylinder is less than that of any Fig. 253. circumscribed prism BCDKLH (fig. 253).

Fig. 258.

THEOREM.

523. The convex surface of a cylinder is equal to the circumference of its base multiplied by its altitude.

Demonstration. Let CA (fig. 258) be the radius of the base of the given cylinder, H its altitude; and let circ. CA be the circumference of a circle whose radius is CA; we say that circ. CA× H will be the convex surface of the cylinder. For, if this proposition be denied, circ. CAx H must be the surface of a cylinder either greater or less; and, in the first place, let us suppose that it is the surface of a less cylinder, of a cylinder, for example, which the radius of the base is CD and the altitude H.

of

Circumscribe about the circle, whose radius is CD, a regular polygon GHIP, the sides of which shall not meet the circumference whose radius is CA; then suppose a right prism, whose altitude is H, and whose base is the polygon GHIP. The convex surface of this prism will be equal to the perimeter of the polygon GHIP multiplied by its altitude H (520); this perimeter is less than the circumference of the circle whose radius is CA; consequently the convex surface of the prism is less than circ. CAx H. But circ. CAx H is, by hypothesis, the convex surface of a cylinder of which CD is the radius of the base, which cylinder is inscribed in the prism; whence the convex surface of the prism would be less than that of the inscribed cylinder. But on the contrary it is greater (522); accordingly the hypothesis with which we set out is absurd; therefore, 1. the circumference of the base of a cylinder multiplied by its altitude cannot be the measure of the convex surface of a less cylinder.

We say, in the second place, that this same product cannot be the measure of the surface of a greater cylinder. For, not to change the figure, let CD be the radius of the base of the given cylinder, and, if it be possible, let circ. CD x H be the convex surface of a cylinder, which with the same altitude has for its base a greater circle, the circle, for example, whose radius is CA. The same construction being supposed as in the first hypothesis, the convex surface of the prism will always be equal to the perimeter of the polygon GHIP, multiplied by the altitude H. But this perimeter is greater than circ. CD; consequently the surface of the prism would be greater than circ. CD × H, which, by hypothesis, is the surface of a cylinder of the same altitude of which CA is the radius of the base. Whence the surface of the prism would be greater than that of the cylinder. But while the prism is inscribed in the cylinder, its surface will be less than that of the cylinder (522); for a still stronger reason is it less when the prism does not extend to the cylinder; consequently the second hypothesis cannot be maintained; therefore, 2. the circumference of the base of a cylinder multiplied by its altitude cannot be the measure of the surface of a greater cylinder.

We conclude then that the convex surface of a cylinder is equal to the circumference of the base multiplied by its altitude.

Fig. 259.

THEOREM.

524. The solidity of a cone is equal to the product of its base by a third part of its altitude.

Demonstration. Let SO (fig. 259) be the altitude of the given cone, AO the radius of the base; representing by surf. AO the surface of the base, we say that the solidity of the cone is equal to surf. AO SO.

1. Let surf. AO × SO be supposed to be the solidity of a greater cone, of a cone, for example, whose altitude is always SO, but of which BO, greater than AO, is the radius of the base.

About the circle, whose radius is AO, circumscribe a regular polygon MNPT, which shall not meet the circumference of which OB is the radius (285); suppose then a pyramid having this polygon for its base and the point S for its vertex. The solidity of this pyramid is equal to the area of the polygon MNPT multiplied by a third of the altitude SO (416). But the polygon is greater than the inscribed circle represented by surf. OA; consequently, the pyramid is greater than

surf. AO × SO,

which, by hypothesis, is the measure of the cone of which S is
the vertex, and OB the radius of the base. But on the contrary
the pyramid is less than the cone, since it is contained in it;
therefore it is impossible that the base of the cone multiplied by
a third of its altitude should be the measure of a greater cone.
2. We say, moreover, that this same product cannot be the
measure of a smaller cone. For, not to change the figure, let
OB be the radius of the base of the given cone, and, if it be
possible, let surf. OB × SO be the solidity of a cone which
has for its altitude SO, and for its base the circle of which AO
is the radius. The same construction being supposed as above,
the pyramid SMNPT will have for its measure the area MNPT
multiplied by SO. But the area MNPT is less than surf. OB;
consequently the pyramid will have a measure less than

surf. OB SO,

and accordingly it would be less than the cone, of which AO is the radius of the base and SO the altitude. But on the contrary the pyramid is greater than the cone, since it contains it; therefore it is impossible that the base of a cone multiplied by a third of its altitude should be the measure of a less cone.

We conclude then, that the solidity of a cone is equal to the product of its base by a third of its altitude.

525. Corollary. A cone is a third of a cylinder of the same base and same altitude; whence it follows,

1. That cones of equal altitudes are to each other as their bases;

2. That cones of equal bases are to each other as their altitudes;

3. That similar cones are as the cubes of the diameters of their bases, or as the cubes of their altitudes.

526. Scholium. Let R be the radius of the base of a cone, H its altitude; the solidity of the cone will be a R2 × Н, oг лRH.

THEOREM.

527. The frustum of a cone ADEB (fig. 260) of which OA, DP, Fig. 260. are the radii of the bases, and PO the altitude, has for its measure

--2

× OP × (AO+ DP + AO × DP).

Demonstration. Let TFGH be a triangular pyramid of the same altitude as the cone SAB, and of which the base FGH is equivalent to the base of the cone. The two bases may be supposed to be placed upon the same plane; then the vertices S, T, will be at equal distances from the plane of the bases; and the plane EPD produced will be in the pyramid the section IKL. We say now, that this section IKL is equivalent to the base DE; for the bases AB, DE, are to each other as the squares of the radii AO, DP (287), or as the squares of the altitudes SO, SP; the triangles FGH, IKL, are to each other as the squares of these same altitudes (407); consequently the circles AB, DE, are to each other as the triangles FGH, IKL. But by hypothesis, the triangle FGH, is equivalent to the circle AB; therefore the triangle IKL is equivalent to the circle DE.

Now the base AB multiplied by SO is the solidity of the cone SAB, and the base FGH multipled by SO is that of the pyramid TFGH; the bases therefore being equivalent, the solidity of the pyramid is equal to that of the cone. For a similar reason the pyramid TIKL is equivalent to the cone SDE; therefore the frustum of the cone ADEB is equivalent to the frustum of the pyramid FGHIKL. But the base FGH, equivalent to the circle of which the radius is AO, has for its measure